Problem: $\vec v = (3,4)$ $-2\vec v= ($
In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $-2 \vec{v}$ : $\begin{aligned} {-2}\vec v = {-2} \cdot (3,4) &= \left({-2} \cdot 3, {-2} \cdot 4\right) \\\\ &= (-6,-8) \end{aligned}$ The answer is $ (-6,-8) $.